Exams Joint Entrance Examination Mathematics Ellipse Question 10 Question 10 of 183 Medium MCQ
Let the line y − x = 1 y-x=1 y − x = 1 intersect the ellipse x 2 2 + y 2 1 = 1 \frac{x^2}{2}+\frac{y^2}{1}=1 2 x 2 + 1 y 2 = 1 at the points A and B . Then the angle made by the line segment AB at the center of the ellipse is :
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y = x + 1 x 2 2 + y 2 = 1 ⇒ x 2 2 + ( x + 1 ) 2 = 1 ⇒ x 2 + 2 x 2 + 4 x = 0 ⇒ x = 0 , − 4 3 A B ( 0 , 1 ) , ( − 4 3 , − 1 3 ) θ = tan − 1 ( − 1 3 − 4 3 ) = tan − 1 ( 1 4 ) ⇒ \angleAOB = π 2 + tan − 1 ( 1 4 ) \begin{gathered} y=x+1\frac{{x}^{2}}{2}+{y}^{2}=1 \\ \Rightarrow \frac{{x}^{2}}{2}+(x+1{)}^{2}=1\Rightarrow {x}^{2}+2{x}^{2}+4x=0 \\ \Rightarrow x=0,\frac{-4}{3} \\ AB \\ (0,1),(\frac{-4}{3},\frac{-1}{3}) \\ \theta ={\tan }^{-1}(\frac{\frac{-1}{3}}{\frac{-4}{3}})={\tan }^{-1}(\frac{1}{4}) \\ \Rightarrow \angleAOB=\frac{\pi }{2}+{\tan }^{-1}(\frac{1}{4}) \end{gathered} y = x + 1 2 x 2 + y 2 = 1 ⇒ 2 x 2 + ( x + 1 ) 2 = 1 ⇒ x 2 + 2 x 2 + 4 x = 0 ⇒ x = 0 , 3 − 4 A B ( 0 , 1 ) , ( 3 − 4 , 3 − 1 ) θ = tan − 1 ( 3 − 4 3 − 1 ) = tan − 1 ( 4 1 ) ⇒ \angleAOB = 2 π + tan − 1 ( 4 1 ) Answer the question to unlock